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Saturday, 31 October

JEE Main 2020 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 8 January 2020, Shift 1, Maths Q57

Question 57 of 75 in this shift, Maths question 7 of 25, Section A.

lim⁡x→0(3x2+27x2+2)1/x2\lim_{x\to 0}\left(\dfrac{3x^2+2}{7x^2+2}\right)^{1/x^2} is equal to :
  1. (1)ee
  2. (2)e2e^2
  3. (3)1e\frac{1}{e}
  4. (4)1e2\frac{1}{e^2}Official answer

Official answer

Option 4

NTA final key (Jan 2020).

Same idea in other shifts

Asked 6× in all
  1. 28 Jul 2022, Shift 1 · Q29lim⁡x→0((x+2cos⁡x)3+2(x+2cos⁡x)2+3sin⁡(x+2cos⁡x)(x+2)3+2(x+2)2+3sin⁡(x+2))100x\lim_{x\to0}\left(\frac{(x+2\cos x)^3+2(x+2\cos x)^2+3\sin(x+2\cos x)}{(x+2)^3+2(x+2)^2+3\sin(x+2)}\right)^{\frac{100}{x}} is equal to…HardNumerical value
  2. 4 Aug 2021, Shift 1 · Q67lim⁡x→2(x3−x2−8x+13)1x3−3x2+4\lim_{x\to 2}\left(x^3-x^2-8x+13\right)^{\frac{1}{x^3-3x^2+4}}MediumSingle correct
  3. 4 Aug 2021, Shift 2 · Q66The value of lim⁡x→2(2−cos⁡(2x2−5x+2))1(x2−4x+4)\lim_{x\to2}\left(2-\cos\left(2x^2-5x+2\right)\right)^{\frac{1}{\left(x^2-4x+4\right)}} is equal toMediumSingle correct
  4. 25 Feb 2021, Shift 1 · Q66lim⁡n→∞(1+1+12+.......+1nn2)n\lim_{n \to \infty}\left(1 + \frac{1 + \frac{1}{2} + ....... + \frac{1}{n}}{n^2}\right)^n is equal to :MediumSingle correct
  5. 2 Sep 2020, Shift 2 · Q59lim⁡x→0(tan⁡(π4+x))1/x\lim_{x \to 0}\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{1/x} is equal to :EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.